Lesson Objectives
• Demonstrate an understanding of arithmetic
• Learn the definition of a variable
• Learn the definition of a term
• Learn the definition of like terms
• Learn the definition of an algebraic expression
• Learn how to simplify an algebraic expression

## Variables, Terms, and Algebraic Expressions

### What is a Variable in Math?

Once we finish pre-algebra and have a good command over arithmetic, it is time to move forward and look at more challenging material. One of the first and most confusing parts of algebra is dealing with the concept of a variable. A variable can be described in many ways, but generally, we describe a variable as a box, container, or placeholder for an unknown value. A variable can be represented with any symbol we would like, however, in algebra the lowercase letters x,y, and z are most common. As an example, suppose a local hardware store pays Jacob $80 per day. In addition to this, Jacob receives tips when helping customers load their vehicles. If we wanted to represent Jacob’s daily earnings, we could use a variable like x: 80 + x The first part, 80, represents the known amount Jacob earns each day from the hardware store. This amount is set and will not change. The second part, x, is a variable or placeholder for the unknown amount earned in tips. If one day Jacob receives$5 in tips for a job well done, we can replace the variable x with the known amount 5 and perform the calculation:
80 + 5 = 85
In that case, Jacob earned $85 for the day. As another example, suppose we arrive at a gas station and the price per gallon for premium unleaded fuel is$6. In most cases, we don't know how many gallons are being purchased, so we can represent our purchase as:
6x
The first part, 6, represents the known price per gallon. The second part, x, represents the unknown number of gallons. Once we know how many gallons are required for our fillup, we can multiply 6 by the known number and calculate our final sale price. Let’s suppose it takes 12 gallons to fill our car on this visit. In this case, we can replace the variable x with the known amount of 12 and perform the calculation:
6 • 12 = 72

### What is a term?

When we move from pre-algebra to algebra, we will be introduced to some new vocabulary and ways of performing certain operations. First and foremost, let’s address the change in how we show the multiplication operation. When we worked in pre-algebra, we often used the "x" symbol to show multiplication between two numbers:
3 x 2 = 6
This shows the operation of 3 multiplied by 2 and the result is 6. Once we get to algebra, things will change. The "x" symbol is no longer used as it may be confused with the variable x. Now we will use a dot placed between the two numbers:
3 • 2 = 6
This form means the same as before, 3 multiplied by 2 is 6. Additionally, we can place numbers or variables next to each other to imply multiplication. If we are working with numbers, we will place one or more of the numbers inside of parentheses:
3(2) = 6
(3)2 = 6
(3)(2) = 6
In each case, we are showing 3 times 2 gives a result of 6. When multiplying a number times a variable, parentheses are not needed:
4x
This form tells us we have 4 multiplied by x. When a number is multiplying a variable, it is referred to as a coefficient. When a number is not multiplying a variable, it is referred to as a constant. A term is a number or a number times one or many variables:
3, 5x, and 7xyz are each an example of a term:
3 » a constant or number
5x » a number (5) multiplied by a variable (x)
7xyz » a number (7) multiplied by three variables: x, y, and z

### What are Algebraic Expressions

An algebraic expression is one or more terms separated by plus (+) or (-) symbols. As an example, suppose we saw:
7xy + 6x - 9
In this case, our algebraic expression is made up of three terms: 7xy, 6x, and 9, and two operations: plus and minus. An algebraic expression will change its value based on the value or values given for the variable or variables. Let's take a look at a few examples.
Example 1: Evaluate each algebraic expression for the given values of the variable
Evaluate: 3x - 1 for x = 1, x = 3, and x = -2
Let's begin with evaluating:
3x - 1 for x = 1
To solve this problem, we plug in a 1 for x
3x - 1 » 3(1) - 1
Now we can evaluate the problem:
3(1) - 1 = 3 - 1 = 2
Evaluating 3x - 1 for x = 1, gives us 2
Let's look at the case where x = 3:
3(3) - 1 = 9 - 1 = 8
Evaluating 3x - 1 for x = 3, gives us 8
Let's look at our final case, where x = -2:
3(-2) - 1 = -6 - 1 = -7
Evaluating 3x - 1 for x = -2, gives us -7

### What are like terms?

When we start simplifying algebraic expressions, we will come across "like terms". Like terms are terms that have the exact same variable parts. This means the terms have the exact same variables and are raised to the exact same powers. Let's take a look at some examples of like terms:
Like Terms
1.7x3x5x
2.4x29x212x2
3.11xy3-3xy3-8xy3
We can see that our first row of the table has: 7x, 3x, and 5x. In each case, our variable is x and our exponent is understood to be 1. In the second row, we see 4x2, 9x2, and 12x2. In each case, our variable is x and our exponent is a 2. In our third and final row, we have 11xy3, -3xy3, and -8xy3. In each case, our variables are x and y. The exponent on each x is understood to be 1. Our exponent on each y is 3. Let's take a look at some examples in which we don't have like terms:
Not Like Terms
1.-4x3xz5xy
2.-x22x4x3
3.14x2y52xy5-x2y3
We can see that our first row of the table has -4x, 3xz, and 5xy. These are not like terms since the variable parts are not the same. Each term must have the exact same variable parts in order to be considered "like terms". In row 2, we have -x2, 2x4, and x3. We can see that the variable x is the same for each, however, the exponents are different. In our last row, we have 14x2y5, 2xy5, and -x2y3. We can see that our variables (x and y) are the same, but the exponents are not the same.

### How to Combine Like Terms

We can combine "like terms" by keeping the variable part(s) the same and performing operations with the coefficients. Let's suppose we had:
3x + 2x
How can we solve this problem? Since 3x and 2x are "like terms", we can add the coefficients (3 + 2) and keep the variable part the same:
3x + 2x = (3 + 2)x = 5x
To think more deeply about this process, suppose we were just adding apples. If we had 3 apples and added 2 apples, apples wouldn't change, we would just add the quantities 3 + 2 = 5. We could then say we had 5 apples. Now suppose we encountered non-like terms. What does that look like visually? Let's suppose we had:
3x + 2y
x and y here are different variables. It would be like combining apples and oranges. They are not the same, so if we had 3 apples and 2 oranges, we can't make that any simpler. We could only say we have 3 apples and 2 oranges. Let's take a look at a few examples.
Example 2: Simplify each by combining like terms
7x2 + 9x2
Since the variable part x2 is the same in each case, we have like terms. To combine like terms, just perform the operation with the coefficients and keep the variable part the same:
7x2 + 9x2 = (7 + 9)x2 = 16x2
Example 3: Simplify each by combining like terms
-4x3y - 6x3y
Since the variable part x3y is the same in each case, we have like terms. To combine like terms, just perform the operation with the coefficients and keep the variable part the same:
-4x3y - 6x3y = (-4 - 6)x3y = -10x3y

### Simplifying an Algebraic Expression

In order to simplify an algebraic expression, we use the distributive property to remove any parentheses and then we combine any like terms. Let's take a look at a few examples.
Example 4: Simplify each algebraic expression
5x + 2(x + 3) - 3
We begin by using our distributive property to remove parentheses:
5x + 2x + 6 - 3
Now we can combine like terms. 5x and 2x are like terms, as are 6, and 3:
5x + 2x = (5 + 2)x = 7x
6 - 3 = 3
7x + 3
7x and 3 are not like terms since 7x has the variable x and 3 has no variable present. Therefore, 7x + 3 is our simplified answer:
5x + 2(x + 3) - 3 = 7x + 3
Example 5: Simplify each algebraic expression
-(x - y) + 3(2x - 4y) - xy
Let's begin by removing parentheses using the distributive property:
-x + y + 6x - 12y - xy
We can now combine like terms. -x and 6x are like terms, as are y and 12y.
-x + 6x = (-1 + 6)x = 5x
y - 12y = -11y = (1 - 12)y = -11y
xy will not be able to be combined with anything else. This term has both variables x and y. It is not like terms with -11y or 5x. Therefore, 5x - 11y - xy is our simplified answer.
-(x - y) + 3(2x - 4y) - xy = 5x - 11y - xy